Quiz Tweek

Preparing your workspace...

Coordination Compounds Set-7 MCQ Online Practice Test | Class 12 | Quiz Tweek

Coordination Compounds Set-7

Practice Important MCQs for Coordination Compounds — Chemistry, Class 12.

Coordination compounds form a core and scoring area in Class 12 Chemistry because they connect electronic structure, oxidation states, geometry, stereochemistry, bonding (ligand effects), and magnetic/optical properties. Board and competitive exams frequently test concepts like CFSE, spin-only moments, ligand-field strength, Jahn–Teller distortions, and stereoisomerism—so mastering this chapter directly improves performance across multiple question types.

Minutes

Questions

1 / -0

Marking

Question Bank Preview

Q1. Determine the oxidation state of cobalt and the number of d‑electrons in the complex $[Co(NH_3)_5Cl]Cl_2$ .

(A)

Oxidation state $+2$ , configuration $d^7$

(B)

Oxidation state $+3$ , configuration $d^6$

(C)

Oxidation state $+1$ , configuration $d^8$

(D)

Oxidation state $+4$ , configuration $d^5$

Q2. How many stereoisomers (including optical isomers) are possible for the complex ion $[Co(en)_2Cl_2]^+$ (en = ethylenediamine, a symmetrical bidentate ligand)?

(A)

Two

(B)

Four

(C)

Five

(D)

Three

Q3. Consider $[Fe(H_2O)_6]^{2+}$ (assume high‑spin) and $[Fe(CN)_6]^{4-}$ (assume low‑spin). Which complex has the larger spin‑only magnetic moment and what is its approximate value?

(A)

$[Fe(H_2O)_6]^{2+}$ ; $\mu_{\text{spin-only}}\approx 4.90\ \text{BM}$

(B)

$[Fe(CN)_6]^{4-}$ ; $\mu_{\text{spin-only}}\approx 4.90\ \text{BM}$

(C)

$[Fe(H_2O)_6]^{2+}$ ; $\mu_{\text{spin-only}}\approx 2.83\ \text{BM}$

(D)

Both complexes have the same spin‑only moment $\approx 0\ \text{BM}$

Q4. Which one of the following complexes is expected to be diamagnetic under normal conditions (consider typical geometry favored by the ligand)?

(A)

$[NiCl_4]^{2-}$

(B)

$[Cu(NH_3)_4]^{2+}$

(C)

$[Ni(CN)_4]^{2-}$

(D)

$[Fe(CN)_6]^{3-}$

Q5. An octahedral $d^6$ complex shows an absorption band at $\lambda=420\ \text{nm}$ attributed to the crystal field splitting $\Delta_o$ . Using the approximation

\Delta_o\ (\text{kJ mol}^{-1})\approx\frac{119627}{\lambda\ (\text{nm})}

and given pairing energy $P=260\ \text{kJ mol}^{-1}$ , predict the spin state and number of unpaired electrons.

(A)

High‑spin $d^6$ , four unpaired electrons

(B)

Low‑spin $d^6$ , zero unpaired electrons

(C)

High‑spin $d^6$ , two unpaired electrons

(D)

Low‑spin $d^6$ , two unpaired electrons

Q6. Calculate the spin-only magnetic moment (in Bohr magneton, BM) for the octahedral complex $[Cr(NH_3)_6]^{3+}$ . Use $\mu_{s.o.}=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

(A)

$1.73\ \text{BM}$

(B)

$4.90\ \text{BM}$

(C)

$3.87\ \text{BM}$

(D)

$0.00\ \text{BM}$

Q7. For the equilibria $M^{2+}+L \xrightleftharpoons[K_1]{} ML^{2+}$ and $ML^{2+}+L \xrightleftharpoons[K_2]{} ML_2^{2+}$ , $K_1=10^{6}\ \mathrm{M^{-1}}$ , $K_2=10^{4}\ \mathrm{M^{-1}}$ . If initial $[M^{2+}]_0=10^{-4}\ \mathrm{M}$ and free ligand $[L]=0.01\ \mathrm{M}$ (large excess), which species will predominate at equilibrium?

(A)

$ML_2^{2+}$ will predominate

(B)

$ML^{2+}$ will predominate

(C)

Free $M^{2+}$ will predominate

(D)

Comparable amounts of $M^{2+}$ and $ML^{2+}$ will be present

Q8. Nickel(II) is $d^8$ . Which of the following best explains why $[Ni(CN)_4]^{2-}$ is square‑planar and diamagnetic whereas $[Ni(H_2O)_6]^{2+}$ is octahedral and paramagnetic?

(A)

Octahedral geometry is always more stable for $d^8$ , so $[Ni(CN)_4]^{2-}$ must be an exception due to steric factors

(B)

CN− is bulky and forces four‑coordinate geometry; H2O is small and allows six coordination

(C)

Square‑planar $d^8$ complexes are always formed regardless of ligand field strength

(D)

CN− is a strong‑field ligand giving large splitting; the square‑planar splitting pattern raises $d_{x^2-y^2}$ and favors pairing in lower orbitals, producing a diamagnetic square‑planar complex, while H2O is weak‑field giving small $\Delta_o$ and a high‑spin octahedral complex

Q9. Assertion (A): The octahedral complex $[Cu(H_2O)_6]^{2+}$ shows a pronounced Jahn–Teller elongation of the axial Cu–O bonds.
Reason (R): $Cu^{2+}$ has electronic configuration $3d^9$ , therefore it possesses one unpaired electron.

(A)

Both A and R are true and R is the correct explanation of A.

(B)

Both A and R are true but R is NOT the correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q10. For $Mn^{3+}$ ( $d^4$ ) in an octahedral complex, $\Delta_o=150\ \mathrm{kJ\,mol^{-1}}$ and pairing energy $P=100\ \mathrm{kJ\,mol^{-1}}$ . Using LFSE and pairing energy, which statement is correct about the preferred spin state and its net electronic stabilization energy?

(A)

Low‑spin is favored; net stabilization $\approx -140\ \mathrm{kJ\,mol^{-1}}$ (i.e., more stable than high‑spin by $\approx 50\ \mathrm{kJ\,mol^{-1}}$ )

(B)

High‑spin is favored; net stabilization $\approx -90\ \mathrm{kJ\,mol^{-1}}$

(C)

Low‑spin is favored; net stabilization $\approx -240\ \mathrm{kJ\,mol^{-1}}$ (pairing energy negligible)

(D)

High‑spin is favored; net stabilization $\approx -150\ \mathrm{kJ\,mol^{-1}}$

Q11. For the octahedral complex $[Mn(H_2O)_6]^{2+}$ determine the number of unpaired electrons and the spin-only magnetic moment (in Bohr magneton, $\mu_B$ ). Use $\mu_{\text{so}}=\sqrt{n(n+2)}\,\mu_B$ where $n$ is the number of unpaired electrons. Assume a high-spin configuration.

(A)

1 unpaired electron; $\mu_{\text{so}}=1.73\ \mu_B$

(B)

3 unpaired electrons; $\mu_{\text{so}}\approx3.87\ \mu_B$

(C)

5 unpaired electrons; $\mu_{\text{so}}\approx5.92\ \mu_B$

(D)

4 unpaired electrons; $\mu_{\text{so}}\approx4.90\ \mu_B$

Q12. Among the following octahedral complexes, which is expected to have the largest crystal field splitting parameter $\Delta_o$ ? (Consider ligand field strength, metal oxidation state and period.)

(A)

$[Ru(NH_3)_6]^{3+}$

(B)

$[Co(NH_3)_6]^{3+}$

(C)

$[Fe(NH_3)_6]^{2+}$

(D)

$[Fe(H_2O)_6]^{2+}$

Q13. How many stereoisomers (including optical isomers) are possible for the complex $[Co(en)_2Cl_2]^+$ (en = ethylenediamine, a bidentate ligand)?

(A)

(B)

(C)

(D)

Q14. For an octahedral complex of a first-row transition metal with a $d^4$ configuration, the crystal field splitting energy is $\Delta_o=220\ \mathrm{kJ\,mol^{-1}}$ and the pairing energy is $P=150\ \mathrm{kJ\,mol^{-1}}$ . Neglect other energetic contributions. Which electronic configuration and number of unpaired electrons will be adopted?

(A)

High-spin $t_{2g}^3e_g^1$ ; 4 unpaired electrons

(B)

Low-spin $t_{2g}^4e_g^0$ ; 2 unpaired electrons

(C)

Low-spin $t_{2g}^3e_g^1$ ; 3 unpaired electrons

(D)

High-spin $t_{2g}^4e_g^0$ ; 0 unpaired electrons

Q15. Consider the ligand exchange reaction $[M(NH_3)_4]^{2+}+2\,\mathrm{en}\longrightarrow[M(en)_2]^{2+}+4\,NH_3$ , where en = ethylenediamine (bidentate). Given the overall formation constants (defined as $\beta=[\text{complex}]/[M][\text{ligand}]^n$ ) $\beta_{NH_3}=10^8$ for $[M(NH_3)_4]^{2+}$ and $\beta_{en}=10^{14}$ for $[M(en)_2]^{2+}$ , the equilibrium constant $K$ for the ligand exchange is closest to:

(A)

$10^{6}$

(B)

$10^{-6}$

(C)

$10^{8}$

(D)

$10^{2}$

...and 5 more challenging questions available in the interactive simulator.

More Quizzes for Class 12

Explore other chapters and subjects, or view all quizzes for this class.

Quiz Tweek

Coordination Compounds Set-7

Question Bank Preview

More Quizzes for Class 12

Chemistry

Mathematics

Physics